Uniqueness of Pexider's functional equation

Let $f:\mathbb{R}\rightarrow\mathbb{R}$, $g:\mathbb{R}\rightarrow\mathbb{R}$, and $h:\mathbb{R}\rightarrow\mathbb{R}$ and consider Pexider's equation,
$$f(x) + g(y) = h(x + y) \qquad \qquad (1)$$
where $f$, $g$ and $h$ are unknown. I assume (for simplicity) that $f$, $g$, and $h$ are twice continuously differentiable. Then, we can find a solution by differentiating with respect to $x$ and $y$,
$$h''(x+y)=0$$
Integrating out, substituting back into $(1)$ and equating coefficients we get,
$$h(z)=c_1z +c_2 + c_3 \qquad f(x)=c_1x +c_2 \qquad g(y)=c_1y +c_3$$
for arbitrary constants $c_1$, $c_2$ and $c_3$.

I have now found a solution to $(1)$. How do I prove that this solution is unique? It seems immediate, but can't quite convince myself.

Answer

Suppose $h$ is twice continuously differentiable. Therefore,

$h''(x+y)=0$ if and only if $h(x+y)=(x+y)k+l$, where $k,l$ are constants （1）

Since $f(x)+g(y)=h(x+y)$, we have
$0=\frac{\partial^2}{\partial x \partial y}(L.H.S.) = h''(x+y)$ （2）

Therefore by (1) and (2) we have $f(x)+g(y)=(x+y) k+l$, and since $x,y$ are arbitrary, $f,g$ have to be linear functions.

So, the solution is unique in this sense.